Chapter 6 The Design of__ feedback Control System

发布于:2021-06-11 03:04:02

Chapter 6 The Design of feedback Control System
6 introduction Ⅰ. Performance specifications Typical time domain indices: Overshoot: Settling time: Raising time: σ% ts tr

Steady state error: Open-loop gain: K

ess

1

Typical frequency domain indices: Cutoff frequency: ? c Phase margin: ? Gain margin: h Peak value :

Mp

The resonant frequency: ? r bandwidth: ?b Typical complex domain indices:

η
ψ
2

Ⅱ Several types of compensator (a) Cascade compensator

R( s )

compensator

process

C (s)

Gc (s)
H (s)

G (s)

(b) Feedback compensator

R( s )

G (s)

C (s)

Gc (s)

H (s)
3

(c) Input compensator(Prefilter)

R( s )

C (s)

Gc (s)

G (s)

H (s)

(d) Compound compensator

R( s )

G p ( s)

Gc (s)
H (s)

G (s)

C (s)

4

Ⅲ Approaches to system design (a). Frequency Method (b). Root Locus Method

(c). Equivalent Structure and Equivalent Transfer Function Method

6.2 Cascade Compensation design using the bode diagram
The compensator Gc (s) can be written

Gc ( s ) ?

K ? ( s ? zi )

M

? (s ? p )
i i ?1
5

i ?1 N

Consider the first-order compensator with transfer function
Gc ( s) ? K (s ? z) ( s ? p)

The phase-lead compensation when z<p.
The phase-lag compensation when z>p.

(a) Phase –lead compensation
C
i

1 1 ? aTs Gc ( s ) ? a 1 ? Ts
v0

?a ? 1?

R1

vi

R2

R1 ? R2 a? R2
R1 R2 T? C R1 ? R2
6

The phase-lead compensation transfer function can be written

Gc ( s ) ?

1 ? aTs 1 ? Ts

?a ? 1?

Bode diagram of the phase-lead network is

dB
? 20

? 2
0

0

?aT ?

?1

? m T ?1
?m

10lg a

20 lg a

?

7

where

?m ?

1 T
?1

a

? m ? sin

a ?1 a ?1

20lg Gc ( j?m ) ? 10lg a
phase- lead design

1 ? aTs Gc ( s ) ? 1 ? Ts

?a ? 1?
8

Example : A feedback control system has a loop transfer function
G( s) H ( s) ? K s( s ? 2)

It is required that the velocity constant equal to 20. The phase-

Margin ? ? 450
Answer:

.

(1) determine value of K
K v ? lim s ? G ( s ) H ( s ) ?
s ?0

K ? 20 2

so

K=40
20 GH ( j? ) ? j? (0.5 j? ? 1)
9

(2) Calculate phase-margin of uncompensated system determine cutoff frequency

40

?c1 (?c1 ) ? 4
2

?1

?c1 ? 6.2rad / s
Calculate phase margin
? ? 1800 ? 900 ? tg ?1 (0.5 ? 6.2) ? 180 ? 450

(3) Calculate the parameter Using Eq.
? m ? sin
?1

a

(Gc ( s ) ?

1 ? aTs 1 ? Ts

?a ? 1?
10

)

a ?1 a ?1

sin

?1

a ?1 ? 45 0 ? 18 0 ? 50 a ?1

a=3.26 (4) Determine the frequency ? m

10lg 3.26 ? 5.1db
20log( 20

?m ?

?m
2

) ? ?5.1

?m ? 8.5rad / s
(5) Evaluate T

?m ?

1 T a

? 8.5
11

T=0.065 (6) Check the rsult

Then compensated loop transfer function is
20(0.21s ? 1) Gc ( s)GH ( s) ? s (0.5s ? 1)(0.065s ? 1)

? (?c ) ? ?900 ? tg ?10.5?c ? tg ?10.065 ?c ? tg ?10.21 ?c ? ?134.90

? ? 1800 ? ? (?c ) ? 45.10

12

Bode diagram

13

Bode diagram

14

Bode diagram

15

Steps: 1. Evaluate the uncompensated system phase margin when error constants are satisfied.

2. Allowing for a small amount of safety , determine the necessary
additional phase lead, ?m .
?1

3. Evaluate a form Eq. ? m ? sin

a ?1 a ?1

4. Evaluate 10loga and determine the frequency ? m uncompensated magnitude curve is equal to –10log a. 5.Evaluate T from Eq. ? m ?
1 T a

,where the

6. Check the result,and repeat the steps if necessary.
16

Example : The block diagram of system is
R
Gc (s)
K s (0.1s ? 1)

C

Determine Gc (s) and K to satisfy the steady-state error≤0.01 due to
r(t)=t, and phase margin≥ 450,cutoff frequency ?c ? 40rad / s

Answer:

set K=100. Uncompensated system has ? ? 1800 ? (?900 ? tg ?13.1) ? 17.90 ?c ? 31rad / s

We set therefore

?c' ? 44rad / s
10log a ? 6
1 T a ? 44

a ? 4 T ? 0.01136
? ? 49.80

The loop transfer function of compensated system is
100(0.04544 s ? 1) Gc ( s)G ( s) ? s(0.01136 s ? 1)(0.1s ? 1)
17

(b) phase-lag compensation R1

vi

R2
C

v0

Vo ( s) R2Cs ? 1 Gc ( s) ? ? Vi ( s) ( R1 ? R2 )Cs ? 1

1 ? Ts 1 ( s ? z ) Gc ( s) ? ? 1 ? aTs a ( s ? p)
a? R1 ? R2 R2

T ? R2C
18

Bode diagram of the phase-lag network is

dB

0

(aT )

?1

?T ??1
? 20lg a

?

? 20
?0

?

?
2

Phase- lag design
1 ? Ts Gc ( s ) ? 1 ? aTs
19

step 1. Obtain the Bode diagram of the uncompensated system with the gain adjusted for the desired error constant. 2. Determine the phase margin of the uncompensated system and , if it is insufficient, proceed with the following steps. 3. Determine the frequency where the phase margin requirement would

be satisfied if the magnitude curve crossed the 0-db line at this ' ? frequency c (Allow for 50 phase lag from the phase-lag
network when determining the new crossover frequency). 4. Place the zero of compensator one decade below the new crossover
' ' 0 ? ? frequency c and thus ensure only 5 of additional phase lag at c

20

5. Calculate a by noting that the attenuation introduced by the phase' lag network is –20loga at ?c .

6.Check the result. Example : A feedback control system has a loop transfer function
G( s) H ( s) ? K s( s ? 2)

It is required that the velocity constant equal to 20. The phaseMargin ? ? 450 answer .

1 ? 6.66 s Gc ( s ) ? 1 ? 66.6 s

a ? 10

21

Cutoff frequency ?c1 ? 6.2rad / s Phase margin

? ? 1800 ? 900 ? tg ?1 (0.5 ? 6.2)
? 180 ? 450 when
? ? 1.5rad / s
? (? ) ? ?90 ? tg ?1 (1.5 / 2) ? ?1270

therefore

set

1 ? 0.15 T

20log a ? 20db

a ? 10
Gc ( s ) ? 1 ? 6.66 s 1 ? 66.6 s
22

The characteristic of phase-lag compensator (1)to increase phase margin. (2) to enhance state-steady accuracy.
? 20loga

?

?
23

(c) Phase Lag- Lead Compensation

(1 ? T1s)(1 ? bT2 s) Gc ( s) ? (1 ? aT1s)(1 ? T2 s)

a ?1
?1

b ? 1 T1 ? bT2

The bode diagram of phase lag-lead network

??T1 ??1
R1

T1

?1

?bT2 ?

?1

T2

vi

C1

R2
C2

v0

(Ta s ? 1)(Tb s ? 1) Gc ( s) ? ?1 (1 ? a Ta s)(1 ? aTb s)
where

Ta ? R1C1

Tb ? R2C2

Tb ? Ta

a ?1

24

Example : A feedback control system has a loop transfer function
G( s) H ( s) ? 126?10? 60 s( s ? 10)(s ? 60)

It is required that the velocity constant equal to 126. The phaseMargin ? ? 300 ,cutoff frequency ?c ? 20rad / s .

Analzing:
Cutoff frequency of the system

?c ? 30.5rad / s ? 20rad / s
When

Phase-lead compensator is not suitable.

? ? 20 rad / s
? ?171.9
0

? (? ) ? ?900 ? tg ?1 (20 / 10) ? tg ?1 (20 / 60)
25

Phase-lag compensator is not suitable. So we should choose phase lag-lead compensator

(Ta s ? 1)(Tb s ? 1) Gc ( s) ? ?1 (1 ? a Ta s)(1 ? aTb s)
The following equations can be written

a ?m ? ? 20 Ta 1 ?2 Tb

?10lg a ? 20lg G( j 20)H ( j 20) ? 0

26

?10lg a ? 20lg 3.15 ? 0
a ? 9.92

Tb ? 0.5
Ta ? 0.1575

(0.1575s ? 1)(0.5s ? 1) Gc ( s) ? (1 ? 0.0159s)(1 ? 4.96s)
check
? ? 1800 ? 900 ? tg ?1 (20/ 20) ? tg ?1 (20/ 60)
?tg ?1 (0.1575? 20) ? tg ?1 (0.5? 20) ? tg ?1 (20 ? 0.0159) ? tg ?1 (4.96? 20)

? 76.2

0

27

28

29

6.3 Cascade Compensation design using the root locus (1) The phase-lead network design The phase-lead network has a transfer function Steps:
s?z Gc ( s ) ? s? p

1. List the system specifications and translate them into a desired root
location for the dominant roots. 2. Sketch the uncompensated root locus,and determine whether the

desired root locations can be realized with an uncompensated system.
3. If a compensator is necessary ,place the zero of the phase-lead

network directly below the desired root location(or to the left of the
first two real poles).
30

4. Determine the pole location so that the total angle at the desired

root location is 1800 and therefore is on the compensated root locus.
5. Evaluate the total system gain at the desired root location and then determine whether the gain satisfied . 6. Repeat the steps if the error constant is not satisfactory.

Example : A feedback system has the open-loop transfer function
K GH ( s ) ? 2 s

It is required the specifications for the system are:
Settling time(2% criterion),Ts ? 4.5s

Percent overshoot for a step input ? 35 %

31

Therefore the damping ratio should be ? ? 0.35

.The settling time

Requirement is
ts ?

4.5 ? 4.5 ??n

Thus we will choose a desired dominant root location as
s ? ?1 ? j 2
900 ? 2 ?1160 ?? p ? ?1800

so

? p ? 380

380

s ?1 Gc ( s ) ? s ? 3.6
K ( s ? 1) Gc ( s)GH ( s) ? 2 s ( s ? 3.6)

K=8.2

Ka ? 8.2 / 3.632 ? 2.3

(2)The phase-lag network design The transfer function of the phase-lag network is
s?z Gc ( s ) ? s? p
K ? ( s ? zi )
i ?1 m

z?

1 T

p?

1 aT
m

a ?1

Assume open-loop transfer function of uncompensated system is
GH ( s ) ?

? (s ? p
j ?1

n

K开 ?

K ? zi
i ?1

j

)

?p
j ?1

n

j

The open-loop transfer function of compensated system is
s?z Gc ( s)GH ( s) ? s? p K ? ( s ? zi )
i ?1 m

? (s ? p )
j ?1 j

n

z K ? K p
' 开

?z
i ?1 n j ?1

m

i

?p

? aK开

j

33

Steps: 1. Obtain the root locus of the uncompensated system. 2. Determine the transient performance specifications for the system

and locate suitable dominant root locations on the uncompensated
root locus that will satisfy the specifications. 3. Calculate the loop gain at the desired root location and the system error constant. 4. Compare the uncompensated error constant with the desired error

constant ,and calculate the necessary increase that result from the
pole-zero ratio of the compensator, a . 5. With the known ratio of the pole-zero combination of the compensator,

determine a suitable location of the pole and zero of the compensator 34

So that the compensated root locus will still pass through the desired

root location.Locate the pole and zero near the origin of the s-plane
in comparison to ??n .

Example :consider the uncompensated system has a open-loop transfer function
GH ( s ) ? K s( s ? 2)

Design a compensation network to satisfy the damping ratio of the
dominant complex roots be 0.45,and velocity constant equal to 20.

35

Set the dominant complex roots
s ? ?1 ? j 2

So

K=5
K开 ? K ? 2 .5 2

K开C ? 2.5 ? a ? 20

therefore

a?

20 ?8 2.5

We can set z=0.1 then p=0.1/8
Therefore the compensated system transfer function is

5( s ? 0.1) Gc ( s)GH ( s) ? s( s ? 2)(s ? 0.0125 )

36

5( s ? 0.1) Gc ( s)GH ( s) ? s( s ? 2)(s ? 0.0125 )

37

6.4 System design using integration networks
Gc ( s ) ? K p ? KI s

Example 10.5:A negative feedback system has a open-loop transfer function K
GH ( s) ?
1

(2s ? 1)(0.5s ? 1)

Determine Gc (s) to maintain zero steady-state error for a step input
and overshoot less that or equal to 10%, settling time(2% criterion) less than or equal to 6 seconds. Answer:
2.08( s ? 0.75) GcGH ( s) ? s(2s ? 1)(0.5s ? 1)
38

Analyzing
?% ? e
??? / 1?? 2

?100% ? 10%

? ? 0.59

4.5

?? n

?6

??n ? 4.5 / 6 ? 0.75
So the determine pole is

? 0.75 ? j1
The angle criterion is

? 1800 ? ?1270 ? 1040 ? 380 ? ? z

? z ? 890

39

KI Gc ( s ) ? K p ? s

KI / KP ? 0.75

So

K1 K P ( s ? 0.75) GcGH ( s) ? s(2s ? 1)(0.5s ? 1)

Due to magnitude criterion
0.752 ? 1 ? 0.52 ? 4 ? 0.6252 ? 0.52 K1K P ? 1

? 2.08

40

6-5 反馈校正

G ( s ) 2 G2? (s) ? 1 ? G2 (s) H (s)
显然,引进H(s)的作用是希望 G2 ? ( s ) 的特性 使整个闭环系统的品质得到改善。
41

反馈校正的几种作用
① 利用反馈改变局部结构、参数 ② 利用反馈削弱非线性因素的影响

③ 反馈可提高对模型扰动的不灵敏性
④ 利用反馈抑制干扰
42

一、利用反馈改变局部结构、参数

① 用位置反馈包围积分环节。
K G(s) ? s
' 2

, H (s) ? K f 1 ,T ? KK f

1 1 G ( s) ? ? K f (Ts ? 1)

使系统的无差度下降,相位滞后减少。
43

② 用速度反馈包围惯性、积分和放大环节。

K G2 ( s) ? s(Ts ? 1)
' 2

, H ( s) ? Kt s

K1 T K G ( s) ? ; T1 ? , K1 ? s(T1 s ? 1) 1 ? KKt 1 ? KKt

可以增加系统的带宽,有利于快速 性的提高。
44

③ 用速度反馈包围一个小阻尼的二阶振荡 环节和放大环节。

K? G2 ( s) ? 2 ,(? ? 1) 2 s ? 2?? n s ? ? n
2

H (s) ? Kt s
2 K ? n G2' ( s ) ? 2 s ? 2(? ? 0.5KK t ? n )? n s ? ? n2

加入速度反馈,增加了阻尼,减弱了小阻尼 环节的不利影响。
45

二、利用反馈削弱非线性因素的影响


G( j? ) G ( j? ) ? 1 ? G( j? ) H ( j? )
'

若满足

| G( j? ) H ( j?) |?? 1

G( j? ) 1 G ( j? ) ? ? G( j? ) H ( j? ) H ( j? )
'
46

三、反馈可提高对模型摄动的不灵敏性

47

四、利用反馈抑制干扰

只要 |1 ? GH |? 1,干扰N的影响就可以得到抑制。 从抑制测量噪声?的角度,要求高频区 | GH |?? 1 。

48

6.6 Systems with a prefilter
R( s )

G p ( s)

Gc (s)

G (s)

C (s)

where Therefore

G (s) ?

1 s

Gc ( s ) ?

K1 s ? K 2 s

?( s ) ?

( K1s ? K2 )Gp (s) s 2 ? K1s ? K2

The specifications require a settling time of 0.5 second and an overshoot of approximately 4%. 4.5 Ts ? ? 0.5 Set ? ? 1/ 2 ?? n

49

?? n ? 9

K1 ? 2?? n ? 18
2 K2 ? ?n ? 162

When Gp (s) ? 1 ,the closed loop transfer function is
?(s) ? 18( s ? 9) s 2 ? 18s ? 162

Set

G p (s) ?

9 s?9

Then we have
?(s) ? 162 s 2 ? 18s ? 162
50

6.7 Compound compensator
G p ( s)

R

C

-

E

G(s)

The closed-loop transfer function is
C ?s ? G ?s ? ?? 1 ? G p ?s ?? R?s ? 1 ? G ?s ?

We required So:

E?s ? ? R?s ? ? C?s ? ? 0
E ?s ? ? R?s ? ? C ?s ? ? 1 ? G?s ?G p ?s ? 1 ? G?s ? R?s ? ? 0

Gp ?s ? ? G?1 ?s ?

51

Set

bm s m ? ? ? ? ? b1 s ? b0 N ( s) G?s ? ? ? n , ?n ? m? n ?1 D( s) s ? a n ?1 s ? ? ? ? ? a1 s ? a0

GP ?s ? ? d2 s2 ? d1s1 ? d0
So the error transfer function is:
E ?s ? s n ? an?1 s n?1 ? ? ? ? ? a1 s ? a0 ? d 2 s 2 ? d1 s1 ? d 0 bm s m ? ? ? ? ? b0 ? R?s ? s n ? ? ? ? ? ?am ? bm ?s m ? ? ? ? ? ?a0 ? b0 ?

?

??

?

The coefficients of constant term, s term and s2 term separate are:

a0 ? d 0b0
a1 ? (d1b0 ? d0b1 )
a2 ? (d0b2 ? d1b1 ? d2b0 )
52

Example :The block diagram of system is

GP (s)
R
K 0.1s ? 1

10 s (0.5s ? 1)

C

(a) Determine the value of K so that overshoot =0; (b) Design GP ( s) ,the steady-state error equal to zero due to r(t)=t. Answer: Breakaway point: d=-0.94 K=0.045

Error transfer transfer is
0.45Gp ( s) ? 0.45 E ( s) ? 1 ? ?( s ) ? 1 ? R( s ) 0.05s3 ? 0.6s 2 ? s ? 0.45

so

GP ( s) ? 2.2s

53

二、对干扰的附加补偿校正

G2 (1 ? G1 Gc ) C ( s) ? N ( s) 1 ? G1 G2
Gc (s) ? ?G (s)
54

?1 1

?1 Gc ( s) ? (T1 s ? 1) K1

55

Gc (s)G1 (s) ? G4 (s) ? 0

G4 ( s) Gc ( s) ? ? G1 ( s)
56

G2 (G1 ? W1 W3 ) 1 C ( s) ? R( s ) ? N ( s) 1 ? G2 (G1 ? W2 W3 ) 1 ? G2 (G1 ? W2 W3 )
57

6.7 Design for deadbeat response A deadbeat response has the following characteristics 1. Steady-state error =0.

2. Fast response

minimum rise time and settling time.

3. 0.1%≤percent overshoot <2%.

4. Percent undershoot <2%.
We consider the transfer function of a closed-loop system,T(s). set
?( s ) ?
2 3 s3 ? ??n s 2 ? ??n s ? ?n 3 ?n

3 Dividing the numerator and denominator by ? n

yields:

?( s) ?

1 s3

?

3 n

??

s2

?

2 n

??

s

? ?1

1 s 3 ? ? s 2 ? ? s ?1

?n

58

Coefficients and response measures of a deadbeat system
Coefficients System Order Percent Overshoot P.O Percent Undershoot P.U 90% Rise time 100% Rise Time Settling Time

?
1.82

?
2.20

?

?

?

Tr 90
3.47

Tr
6.58

Ts
4.82

2nd

0.10%

0.00%

3rd

1.90

1.65%

1.36%

3.48

4.32

4.04

4th

2.20

3.50

2.80

0.89%

0.95%

4.16

5.29

4.81

5th

2.70

4.90

5.40

3.40

1.29%

0.37%

4.84

5.73

5.43

6th

3.15

6.50

8.70

7.55

4.05

1.63%

0.94%

5.49

6.31

6.04

59

Example: Design of a system with a deadbeat response
R( s )

G p ( s)

Gc (s)

G (s)

C (s)

where

K G( s) ? s( s ? 1)

(s ? z) Gc ( s) ? ( s ? p)

G p ( s) ?

z (s ? z)

The closed-loop transfer function is
?( s) ? Kz s 3 ? (1 ? p) s 2 ? ( K ? p) s ? Kz

If we select a settling time of 2 seconds ,then ?nTs ? 4.04 ,and thus

?n ? 2.02
60

The required closed-loop system has the characteristic equation
2 3 q(s) ? s3 ? ??n s 2 ? ??n s ? ?n

? s 3 ? 3.84s 2 ? 8.98s ? 8.24
so
p ? 2.84

z ? 1.34

K ? 6.14

61


相关推荐

最新更新

猜你喜欢